Consider 10 observations x_1, x_2, ldots, x_{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $n=10$ observations $x_1, x_2, \ldots, x_{10}$.From $\sum_{i=1}^{10}(x_i-\alpha)=2$,we have $\sum x_i - 10\alpha = 2$.Given mean $\mu = \fra

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given $n=10$ observations $x_1, x_2, \ldots, x_{10}$.From $\sum_{i=1}^{10}(x_i-\alpha)=2$,we have $\sum x_i - 10\alpha = 2$.Given mean $\mu = \frac{\sum x_i}{10} = \frac{6}{5}$,so $\sum x_i = 12$.Substituting $\sum x_i = 12$ into the first equation: $12 - 10\alpha = 2$ $\Rightarrow 10\alpha = 10$ $\Rightarrow \alpha = 1$.Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2 = \frac{84}{25}$.We know $\sum (x_i - \beta)^2 = \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 40$.From $\sigma^2 = \frac{\sum x_i^2}{10} - (\frac{6}{5})^2 = \frac{84}{25}$,we get $\frac{\sum x_i^2}{10} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25} = \frac{24}{5}$,so $\sum x_i^2 = 48$.Substituting $\sum x_i^2 = 48$ and $\sum x_i = 12$ into the equation $48 - 2\beta(12) + 10\beta^2 = 40$:$10\beta^2 - 24\beta + 8 = 0 \Rightarrow 5\beta^2 - 12\beta + 4 = 0$.Solving the quadratic: $(5\beta - 2)(\beta - 2) = 0$.Since $\beta$ is a positive integer,$\beta = 2$.Thus,$\frac{\beta}{\alpha} = \frac{2}{1} = 2$.

Similar Questions

If the variance of $10$ natural numbers $1, 1, 1, \dots, 1, k$ is less than $10$,then the maximum possible value of $k$ is ...... .

The coefficient of variation of the first $5$ prime numbers is

The variance of the first $20$ natural numbers is

If the standard deviation of the numbers $2, 3, a$,and $11$ is $3.5$,then which of the following is true?

$A$ student scored the following marks in five tests: $45, 54, 41, 57, 43$. His score is not known for the sixth test. If the mean score is $48$ in the six tests,then the standard deviation of the marks in the six tests is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module